SOLUTION: Can someone help me with this problem? Thank you. y=e^x; t(x,y)=(x-2, y+3) Write the equation of the image under the given transformation in simplest form. Then transform three

Algebra ->  Graphs -> SOLUTION: Can someone help me with this problem? Thank you. y=e^x; t(x,y)=(x-2, y+3) Write the equation of the image under the given transformation in simplest form. Then transform three       Log On


   



Question 1100876: Can someone help me with this problem? Thank you.
y=e^x; t(x,y)=(x-2, y+3)
Write the equation of the image under the given transformation in simplest form. Then transform three points from the pre-image to their image points under the given mapping.

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
y=e%5Ex t%28x%2Cy%29=%22%28+x-2+%2C+y%2B3%29%22 .
Not all the language of math is universal or standardized, but
I interpret the problem as every point in the graph is moving.
-2 units to the right (meaning 2 units to the left),
and 3 units up,
and we achieve that by placing x%2B2 where x goes,
and y-3 where y goes,in the expression y=e%5Ex .
If my interpretation is the one used in your class,
The plots and equation of the image follow.
Let us take 3 points from the graph of y=e%5Ex .
Let us tabulate the values for points A, B and C with x=-2, x=0 and x=1
.
The plotted points and the graph of the function look like this


The new tabulation for image points %22A+%27%22 ,%22B+%27%22 , and %22C+%27%22 ,
adding -2 to each x value, and 3 to each y value, is
.

The plot for old (pre-image) and (image) new points/function can be graphed like this:
.
The equation for the red graph would be
y-3=e%5E%28x%2B2%29 , but y=e%5E%28x%2B2%29%2B3 could be considered to be in "simplest form".